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I’ve run across this problem whilst studying for a comprehensive exam, and would appreciate any hints on a solution to it.

Let $C:=C\left(\mathbb{R},\mathbb{R}\right)$ denote the space of continuous real-valued functions on the real line. Define a map $\phi:\left(C\times C\right)\rightarrow C$ by $\phi\left(f,g\right):=f\circ g$. Assuming that $C$ is equipped with the topology of pointwise convergence, and $C\times C$ is equipped with the product topology, determine whether or not $\phi$ is a continuous function.

My instinct is to say no, but I don’t have any justification for this. Again, I’d appreciate any suggestions.

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HINT: Let $z$ be the zero function; of course $\varphi(z,z)=z$. For $f\in C$, finite $F\subseteq\Bbb R$, and $\epsilon>0$ let

$$B(f,F,\epsilon)=\{g\in C:|g(x)-f(x)|<\epsilon\text{ for all }x\in F\}\;;$$

the family of all such sets $B(f,F,\epsilon)$ is a base for the topology on $C$. To show that $\varphi$ is not continuous, it suffices to show that if $F_0,F_1\subseteq\Bbb R$ are finite and $\epsilon_0,\epsilon_1>0$, there are $f_0\in B(z,F_0,\epsilon_0)$ and $f_1\in B(z,F_1,\epsilon_1)$ such that $\varphi(f_0,f_1)\notin B(z,\{0\},1)$. In other words, we must show that there are $f_0,f_1\in C$ such that $|f_0(x)|<\epsilon_0$ for all $x\in F_0$, $|f_1(x)|<\epsilon_1$ for all $x\in F_1$, and $\left|f_0\big(f_1(0)\big)\right|\ge 1$. Can you see how to do this? The key idea is in the spoiler-protected block below, if you get completely stuck.

Choose $f_1\in B(z,F_1,\epsilon_1)$ so that $f_1(0)\notin F_0$; this is always possible. Then choose $f_0\in B(z,F_0,\epsilon_0)$ so that $f_0\big(f_1(0)\big)=1$; again, this is always possible.

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  • $\begingroup$ Thank you; that is very helpful. I was having a hard time visualizing what is essentially a four-dimensional product space. $\endgroup$ – anonymous Dec 7 '13 at 22:54
  • $\begingroup$ @anonymous: You’re welcome. $\endgroup$ – Brian M. Scott Dec 7 '13 at 23:02
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Hint: Given two topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$, and a function $f:X\rightarrow Y$, we say that $f$ is continious with respect to the topologies $\tau_X$ and $\tau_Y$ if: $$ f^{-1}(U) \in \tau_x \forall \ U\in\tau_Y $$

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  • $\begingroup$ This is not useful. Anyone studying for a comprehensive exam in topology will already be well aware of this very elementary fact. $\endgroup$ – Brian M. Scott Dec 7 '13 at 21:35

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